Question

Consider an infinite G.P with first term $a$ and common ratio $r$, its sum is $4$ and the second term is $\frac{3}{4}$, then

• A. $a=\frac{4}{7},r=\frac{3}{7}$
• B. $a=\frac{3}{2},r=\frac{1}{2}$
• C. $a=2,r=\frac{3}{8}$
• D. $a=3,r=\frac{1}{4}$

Answer 1

The correct option is D

$a=3,r=\frac{1}{4}$

Explanation for the correct options:

Step1: Finding the value of $a$

Given first term $=a$

Common ratio $=r$

Second term, $ar=\frac{3}{4}$

$\Rightarrow$ $r=\frac{3}{4a}$

Sum of infinite GP, $S=4$

$\Rightarrow$ $\frac{a}{1-r}=4$

$\Rightarrow$ $\frac{4a^2}{4a-3}$$=4$

$\Rightarrow$ $(a-3)(a-1)=0$

So $a=3or1$

Step 2: Finding the value of $r$

When $\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{,}\mathbf{}\mathit{r}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}$

When $a=1,r=\frac{3}{4}$

Hence, Option(D) is correct.